Problem: If $x \diamond y = x^{2}-4y^{2}$ and $x \boxdot y = xy+2x-y$, find $4 \diamond (1 \boxdot 4)$.
Explanation: First, find $1 \boxdot 4$ $ 1 \boxdot 4 = 4+(2)(1)-4$ $ \hphantom{1 \boxdot 4} = 2$ Now, find $4 \diamond 2$ $ 4 \diamond 2 = 4^{2}-4(2^{2})$ $ \hphantom{4 \diamond 2} = 0$.